python 3.x需要帮助将一个文本文件拆分为4个字典

Question

有人能帮我解决我遇到的问题吗？我正在试着写一些代码，可以让用户选择特定的文件，然后将其与当前的技能水平进行比较，每个文件都有一个简单、中等、困难和精英的部分，我想把它们放在单独的字典中，我可以让它打印整个文件的正确信息，但我想不出如何将它分成4个单独的字典

# shows all osrs diarys 
def diary_selection():
    diary_options = {
    0 : 'ardougne', 1 : 'desert', 2 : 'falador', 3 : 'fremennik', 4 : 'kandarin',
    5 : 'lumbridge', 6 : 'morytania', 7 : 'varrock', 8 : 'western', 9 : 'wilderness'
    }
    print(diary_options)

# if not a correct number gives prompt to retry  
    while True:
        try:
            d_sel = int(input ("\nType in the number next to the diary: "))
            diary_select = d_sel

            if not (0 < diary_select > 9):
                print('option is valid')
                break
            else:
                print(" invalid option.")
                diary_selection()

        except ValueError:
            print(" invalid option, numbers only, please try again.")


# gets the name of the diary acording to what number was pressed
    current_diary = diary_options.get(diary_select)

#creats a filename for spicific diary        
    diary_file_name = str((current_diary + "_diary.txt"))
    print (diary_file_name,"\n")

#searches for file name in folder ./diary_requirements     
    f = open("./diary_requirements/"+diary_file_name,"r")
    file_contents = f.read()


    return file_contents

我试图操作的文件被组织在一个txt文件中，格式为简单、中等、困难、精英级别。

easy_levels = {
"Attack" : 0
, "Defense" : 0
, "Strength" : 0
, "Hitpoints": 0
, "Range" : 30
, "Prayer" : 0
, "Magic" : 0
, "Cooking" : 0
, "Woodcutting" : 0
, "Fletching" : 20
, "Fishing" : 0
, "Firemaking" : 0
, "Crafting" : 0
, "Smithing" : 0
, "Mining" : 15
, "Herblore" : 0
, "Agility" : 0
, "Thieving" : 0
, "Slayer" : 0
, "Farming" : 0
, "Runecrafting" : 0
, "Hunting": 9
, "Construction" : 0
,
}
medium_levels = {
"Attack" : 0
, "Defense" : 0
, "Strength" : 0
, "Hitpoints": 0
, "Range" : 30
, "Prayer" : 0
, "Magic" : 0
, "Cooking" : 42
, "Woodcutting" : 35
, "Fletching" : 5
, "Fishing" : 46
, "Firemaking" : 35
, "Crafting" : 0
, "Smithing" : 0
, "Mining" : 40
, "Herblore" : 0
, "Agility" : 37
, "Thieving" : 0
, "Slayer" : 0
, "Farming" : 0
, "Runecrafting" : 0
, "Hunting": 31
, "Construction" : 0
,
}

Answer 1

我猜你唯一的困难就是如何根据你所描述的结构填写四本字典。

如果您确定这些文件不会被除您以外的任何人更改，并且您可以使用不安全和肮脏的代码，您可以这样做：

exec(file_contents)

这样做的目的是，由于您的文件结构的内容已经是有效的python，所以它在被调用的作用域中执行它。因此，在执行它之后，您可以访问调用它的作用域中的变量easy_levels, medium_levels, hard_levels, elite_levels。请注意，这假设您在尝试访问的任何日记中都正确定义了这些变量，如果每个日记定义的变量可能会发生变化，您应该使用更安全的方法(或丑陋的快速访问locals())。

Answer 2

试试这个：

import os
import imp
from pprint import pprint

# shows all osrs diarys 
def diary_selection():
    diary_options = {
    0 : 'ardougne', 1 : 'desert', 2 : 'falador', 3 : 'fremennik', 4 : 'kandarin',
    5 : 'lumbridge', 6 : 'morytania', 7 : 'varrock', 8 : 'western', 9 : 'wilderness'
    }
    print(diary_options)

    # if not a correct number gives prompt to retry  
    while True:
        try:
            d_sel = int(input ("\nType in the number next to the diary: "))
            diary_select = d_sel

            if diary_select in diary_options:
                print('option is valid')
                break
            else:
                print(" invalid option.")
                #diary_selection()

        except ValueError:
            print(" invalid option, numbers only, please try again.")


    # gets the name of the diary acording to what number was pressed
    current_diary = diary_options.get(diary_select)

    #creats a filename for spicific diary        
    diary_file_name = str((current_diary + "_diary.txt"))
    print (diary_file_name,"\n")

    #searches for file name in folder ./diary_requirements     
    #f = open("./diary_requirements/"+diary_file_name,"r")
    #file_contents = f.read()
    #return file_contents

    foo = imp.load_source('userInfo', os.getcwd() + '/diary_requirements/' + diary_file_name)
    print('{}\nEasy levels\n{}'.format('-'*40, '-'*40))
    pprint(foo.easy_levels)
    print('{}\nMediyum levels\n{}'.format('-'*40, '-'*40))
    pprint(foo.medium_levels)

diary_selection()

输出(python )：

{0: 'ardougne', 1: 'desert', 2: 'falador', 3: 'fremennik', 4: 'kandarin', 5: 'lumbridge', 6: 'morytania', 7: 'varrock', 8: 'western', 9: 'wilderness'}

Type in the number next to the diary: 6
option is valid
morytania_diary.txt 

----------------------------------------
Easy levels
----------------------------------------
{'Agility': 0,
 'Attack': 0,
 'Construction': 0,
 'Cooking': 0,
 'Crafting': 0,
 'Defense': 0,
 'Farming': 0,
 'Firemaking': 0,
 'Fishing': 0,
 'Fletching': 20,
 'Herblore': 0,
 'Hitpoints': 0,
 'Hunting': 9,
 'Magic': 0,
 'Mining': 15,
 'Prayer': 0,
 'Range': 30,
 'Runecrafting': 0,
 'Slayer': 0,
 'Smithing': 0,
 'Strength': 0,
 'Thieving': 0,
 'Woodcutting': 0}
----------------------------------------
Mediyum levels
----------------------------------------
{'Agility': 37,
 'Attack': 0,
 'Construction': 0,
 'Cooking': 42,
 'Crafting': 0,
 'Defense': 0,
 'Farming': 0,
 'Firemaking': 35,
 'Fishing': 46,
 'Fletching': 5,
 'Herblore': 0,
 'Hitpoints': 0,
 'Hunting': 31,
 'Magic': 0,
 'Mining': 40,
 'Prayer': 0,
 'Range': 30,
 'Runecrafting': 0,
 'Slayer': 0,
 'Smithing': 0,
 'Strength': 0,
 'Thieving': 0,
 'Woodcutting': 35}